## Algebra 2 (1st Edition)

Using the given identities: $\sec x\tan x-\sin x=\frac{1}{\cos x}\frac{\sin x}{\cos x}-\frac{\sin x\cos^2x}{\cos^2x}=\frac{\sin x}{\cos^2 x}-\frac{\sin x\cos^2x}{\cos^2x}=\frac{\sin x(1-\cos^2 x)}{\cos^2 x}=\frac{\sin^3 x}{\cos^2 x}=\sin x\tan^2 x$ Thus we have shown what we have to.